\section{Nisnevich Topology}
\noindent In this section we study the completely decomposed topology developed by Yevsey Nisnevich, now, called after him. 

\tcr{Importance of Nisnevich topology}

\begin{definition}[Completely Decomposed Morphism]
Let $f:Y\rightarrow X$ be a morphism of schemes, $x\in X$, we call $f$ completely decomposed at $x$ iff there exist $y$ over $x$ 
such that the canonical induced morphism on residue fields is an isomorphism:
\[{f^{\#}_{\rf(y)}}:\rf(x)\longrightarrow \rf(y)\]
\end{definition}
\begin{remark}
\em{Such points $y$ are in on one-to-one correspondence to the lifts
\[\xymatrix{&&Y\ar[d]^f\\
\Spec\ \rf(x)\ar@{^(->}[rr]_{i_x}\ar@{-->}[rru]^l&&X
}\]
}
where $\Imm(i_x)=\{x\}$.
\end{remark}
\begin{proof}
For the above lift diagram, let $y=\Imm(l)$, then $f(y)=x$, and we have the following commutative diagram
\[\xymatrix{\bcO_{X,x}\ar[d]\ar[rr]^{f_y^{\#}}&&\bcO_{Y,y}\ar[d]\ar[dll]|{l_{x}^{\#}}\\
\rf(x)\ar@{-->}[rr]&&\rf(y)
}\]
Assume for the sake of contradiction that $\exists b\in \bcO_{Y,y}$ such that $l_x^{\#}$ mapped to a unit, then the induced 
morphism on the residue fields is zero...\tcr{get to the contradiction}. Hence, $l_x^{\#}$ maps $\gom_{Y,y}$  to zero, i.e. there 
is a unique morphism $\rf(y)\rightarrow\rf(x)$ such that makes the following diagram commute:

\[
\xymatrix{\bcO_{Y,y}\ar[r]\ar[d]&\rf(x)\\
\rf(y)\ar@{-->}[ur]&}
\]
\tcb{using the commutativity of the previous two diagrams}, we see that $\rf(x)\cong \rf(y)$.\\

\noindent On the other hand let $y\in Y$ such that ${f^{\#}_{\rf(y)}}:\rf(x)\longrightarrow \rf(y)$ is an isomorphism, then $l$ is 
defined on topological spaces by $i([(0)])=y$,and on sheaves it is induced by the canonical morphism $\bcO_{Y,y}\rightarrow \rf
(y)\cong\rf(x)$.
\end{proof}

\tcr{What is that is completely decomposed?? In what sense?}
\tcr{bijection with lefts} connection to model category.
\begin{lemma}\cite{hoyois}\label{Nis_BaseChange_OneMorphism}
Completely decomposed morphisms are stable under base change.
\end{lemma}
\begin{proof}
Let $f:Y\rightarrow X$ be completely decomposed at $x$, $g:X'\rightarrow X$ a morphism f schemes with $x'\in X':g(x')=x$. Consider 
the induced morphism $g_{x'}:\Spec \rf(x')\rightarrow \Spec \rf(x)$, and the commutative diagram:
\[\xymatrix{
&X'\times_X Y\ar[rr]\ar[d]&&Y\ar[d]^f\\
&X'\ar[rr]_g&&X\\
\Spec \rf(x')\ar@{^(->}[ur]_{i_{x'}}\ar[rr]_{g_{x'}}&&\Spec \rf(x)\ar@{^(->}[ur]_{i_x}&
}\]
Since $f$ is completely decomposed at $x$, we have a lift:
\[\xymatrix{&&Y\ar[d]^f\\
\Spec\ \rf(x)\ar@{^(->}[rr]_{i_x}\ar@{-->}[rru]^l&&X
}\]
Composing with $g_{x'}$, and from the commutativity of the below cell of the previous diagram, we have the commutative diagram:
\[\xymatrix{
\Spec \rf(x)\ar[rrr]^l\ar[d]_{g_{x'}}&&&Y\ar[d]^f\\
\Spec \rf(x')\ar[rr]_{i_{x'}}&&X'\ar[r]_g&X
}\]
Then by the universal property of the pull-back we have the commutative diagram (lift)
\[\xymatrix{&&X'\times_XY\ar[d]\\
\Spec\ \rf(x')\ar@{^(->}[rr]_{i_{x'}}\ar@{-->}[rru]^l&&X'
}.\]
Hence, $X'\times_XY\rightarrow X'$ is totally decomposed at $x'$.


then the induced pull-back projection $X'\times_X Y\rightarrow X$ is also completely decomposed at $x$,tha
\end{proof}
\noindent If $f:Y\rightarrow X$ is locally of finite representation and completely decomposed at $x\in X$, then $f$ is unramified 
at $y\in Y$, for each $y\in Y$, such that $x=f(y)$ \tcr{. Is it also flat?}. However, the contrary is not correct, as illustrated 
in the below example.
\begin{example} 



\end{example}
An example when $\kk=\bar{\kk}$
\begin{example}
The morphism $f:\Spec \CC\rightarrow\Spec \RR$, induced by the embedding $\RR\hookrightarrow \CC$, is an \'{e}tale morphism, but 
not completely decomposed.
\end{example}
\tcr{Find more interesting example}





\noindent In this section, let $S$ be a \tcr{Noetherian scheme of finite Krull dimension}, $\Sm/S$ the category of \tcr{smooth 
schemes of finite type} over $S$.
\begin{definition}[Nisnevich Covering]\label{NisCov}
We define Nisnevich covering to be a \tcr{finite} family $\bcU=\{f_i:X_i\rightarrow X\}_{i\in I}$ of \'etale morphisms in $\Sm/S$, 
such that for every point $x\in X$, there exist $i\in I$ where $f_i$ is completely decomposed a $x$.
\end{definition}
\tcr{what is the importance of finiteness!?}
\noindent \tcb{It actually covers $X$, in sense of surjectivity.}\\
\tcb{Give example of one surjective \'etale morphism that does not form Nisnevich covering.}\\
\tcb{Example of \'etale covering that is not surjective even collectively, if there is one!}\\
\begin{lemma}\label{Nis_Coprod_Mor}
Let $\bcU=\{f_i:X_i\rightarrow X\}_{i\in I}$ be a finite family of scheme morphisms, then $\bcU$ is a Nisnevich covering iff the 
canonical morphism $\displaystyle\coprod_{i\in I} X_i\rightarrow X$ is completely decomposed at every $x\in X$.
\end{lemma}
\begin{proof}
Let $\bcU$ be a Nisnevch covering, $x\in X$, then $\exists i\in I$ such that we have the lift:
\[\xymatrix{&&X_i\ar[d]^{f_i}\\
\Spec\ \rf(x)\ar@{^(->}[rr]_{i_x}\ar@{-->}[rru]^l&&X
}\]
Composing with the commutative diagram:
\[
\xymatrix{
X_i\ar[d]_{f_i}\ar[rr]^{i_{X_i}}&&\displaystyle\coprod_{i\in I} X_i\ar[lld]\\
X}
\]
we have the lift:
\[\xymatrix{&&\displaystyle\coprod_{i\in I} X_i\ar[d]\\
\Spec\ \rf(x)\ar@{^(->}[rr]_{i_x}\ar@{-->}[rru]^{i_{X_i}\circ l}&&X
}\]
$\displaystyle\coprod_{i\in I} X_i\rightarrow X$ is completely decomposed at every $x$.\\

\noindent Let $\displaystyle\coprod_{i\in I} X_i\rightarrow X$ be completely decomposed at every $x\in X$. \tcb{Using the 
definition of $\displaystyle\coprod_{i\in I} X_i$ type the rest} \tcr{(Finitness is needed here.)}
\end{proof}
Nisnevich covering could have been defined for arbitrary family, not necessary finite, but then the previous lemma, does not hold 
(give an example).
\begin{remark}\label{Finit_Distribution} Let $I$ be a finite set, $\{f_i:X_i\rightarrow X\}$ and $g:Y\rightarrow X$ in $\Sch/S$, 
then
\[ (\displaystyle\coprod_{i\in I} X_i)\times_X Y\cong \displaystyle\coprod_{i\in I} X_i\times_X Y\]
\end{remark}
\begin{proof}
\tcr{Type, based on tensor product of rings is distributed over direct product of rings.}
\end{proof}
Note that this fact does not necessary hold for $I$ infinite. \url{http://math.stackexchange.com/questions/118790/examples-
proving-why-the-tensor-product-does-not-distribute-over-direct-products}
\begin{corollary}A direct result of \ref{Nis_Coprod_Mor}, \ref{Nis_BaseChange_OneMorphism} and \ref{Finit_Distribution} is that 
Nisnevich covering is stable under base change.
\end{corollary}

\begin{lemma}
\noindent The collection of morphisms in $\Sm/S$ that satisfies the condition of definition \ref{NisCov} forms a pre-topology on $
\Sm/S$, in the sense of Grothendieck.
\end{lemma}
\begin{proof}

\end{proof}
This lemma should come before the definition.

\noindent \begin{definition}[Nisnevich Sheaf] Let $\bcA$ be a category with all products, particularly it has a terminal $\ast$, an 
$\bcA$-valued pre-sheaf $F$ on $\Sm/S$, is called Nisnevich sheaf iff it is a sheaf with respect to Nisnevich topology on $\Sm/S$ 
that satisfies $F(\emptyset)=\ast$.
\end{definition}

\tcr{In this section, it is not necessary smooth}
\begin{proposition}
Let $f:Y\rightarrow X$ be an \'etale morphism and $x\in X$, consider \tcr{$\Spec \bcO_{X,x}^h\nhookrightarrow X$ }, then the 
following are equivalent:
\begin{itemize}
\item $Y\rightarrow X$ is completely decomposed at $x$.
\item the morphism $Y\times_X \Spec \bcO_{X,x}^h \rightarrow \Spec \bcO_{X,x}^h$ has sections.
\end{itemize}
\end{proposition}
\begin{proof}
\tcb{Hoyois and Milne} \tcr{using the the theorem above, but where do we used the fact that the morphisms a \'etale. And what is 
the impact of henselization}
\end{proof}
\begin{question}
\tcr{Is there a completely decomposed morphism at some point of the base, that is not \'etale around it?}
\end{question}

Next, we recall the concept of elementary distinguished square, that works as a generator of the Nisnevich topology, as they encode 
all the information about Nisnevich topology. That we will say that each pre-sheaf on $\Sch/S$  is determined to be sheaf or not, 
based on its behaviour on the distinguished square
\begin{definition}\label{ElemDisSq}
We call the Cartesian square
\[\xymatrix{
U\times_X Y\ar[r]\ar[d]&Y\ar[d]^p \\
U\ar@{^(->}[r]_i& X
}
\]
in $\Sch/S$ an elementary distinguished square when $p$ is \'etale, $i$ is an open embedding, and the projection $p_U:(X-i(U))_
{red}\times_X Y \rightarrow (X-i(U))_{red}$ is an isomorphism:
\[
\xymatrix{
(X-i(U))_{red}\times_X Y\ar[rr]\ar[d]_{p_U}&&Y\ar[d]^p\\
(X-i(U))_{red}\ar@{^(->}[rr]|{/}&&X
}
\]

\end{definition}
\tcr{here we follow Hoyois}

\begin{definition}[Splitting Sequence]
Let $\bcU=\{f_i:X_i\rightarrow X\}_{i\in I}$ be a family of morphisms in $\Sch/S$, we call the sequence of closed sub-schemes:
\[
\emptyset=Z_{n+1}\nhookrightarrow Z_n \nhookrightarrow Z_{n-1} \nhookrightarrow ... \nhookrightarrow Z_0=X
\]
a splitting sequence of $\bcU$ if all below projections $p_{i,j}:(Z_j-Z_{j+1})\times_X X_i\rightarrow Z_j-Z_{j+1}$ have \hyperref
[Section]{$S\!-\!$sections} $s_{i,j}:Z_j-Z_{j+1}\rightarrow (Z_j-Z_{j+1})\times_X X_i$ for $i\in I, 0\leq j\leq n$:
\[
\xymatrix{
(Z_j-Z_{j+1})\times_X X_i\ar@{^(->}[rr]\ar[d]^{p_{i,j}}&&Z_j\times_X X_i\ar@{^(->}[rr]\ar[d]&&X_i\ar[d]^{f_i}\\
Z_j-Z_{j+1}\ar@{^(->}[rr]|{\bigcirc}\ar@/^/@{-->}[u]^{s_{i,j}}&&Z_j\ar@{^(->}[rr]|{/}&&X
}
\]
\end{definition}
\tcr{The importance of having a splitting sequence?}
\begin{definition}[Rational section]
Let $f:Y\rightarrow X$ be an morphism of schemes, we say that $f$ has a rational section, if there is an open sub-scheme $U$ in $X
$, where $|U|$ dense in $|X|$ such that projection $p:U\times_X Y\rightarrow U$ has a \hyperref[Section]{section}.
\end{definition}
\begin{lemma}
Let $f:Y\rightarrow X$ be an morphism of schemes then $f$ has a rational section iff every $f_i:Y_i\rightarrow X$ has a rational 
section for every $i\in I$, where $\{Y_i\}_{i\in I}$ are the irreducible components of $Y$.
\end{lemma}
\begin{proof}
Assume $f$ has a rational section $s$ over $\xymatrix{U\ar@{^(->}[r]|{\circ}&X}$, $p$ is the projection $p:Y\times_X U\rightarrow 
U$, then $p\circ s=id_U$. Let  $\xymatrix{Y_i\ar@{^(->}[r]|{\circ}&Y}$ be an irreducible component of $Y$, we have the Cartesian 
diagram:
\[\xymatrix{
&Y\ar[r]^f&X\\
Y_i\ar@{^(->}[ur]|{\circ}&Y\times_X U\ar[r]^p\ar@{^(->}[u]|{\circ}&U\ar@{^(->}[u]|{\circ}\ar@/^/[l]^s\\
Y_i\times_X U\ar@{^(->}[u]|{\circ}\ar@{^(->}[ru]|{\circ}_{p_i}&&
}
.\]
Then, consider the Cartesian diagram:

\[\xymatrix{
&Y\times_X U&U\ar[l]_s\\
Y_i\times_X U\ar@{^(->}[ru]|{\circ}_{p_i}&V\ar[l]^{\pi_s}\ar@{^(->}[ur]|{\circ}&
}
\]
where $V:=(Y_i\times_X U)\times_{(Y\times_X U)} U$. \tcr{We, notice that $|V|$ is dense in $|X|$ since $Y_i$ is irreducible.} Then, 
we consider the cartesian diagram:
\[\xymatrix{
&Y\ar[r]^f&X\\
Y_i\ar@{^(->}[ur]|{\circ}&Y\times_X V\ar[r]^{p'}\ar@{^(->}[u]|{\circ}&V\ar@{^(->}[u]|{\circ}\\
Y_i\times_X V\ar@{^(->}[u]|{\circ}\ar@{^(->}[ru]|{\circ}_{p'_i}&&
}
.\]
Notice, that $Y_i\times_X V\cong (Y_i\times_X U)\times_U V$, then considering the identity $id_V$, and the projection $\pi_s$ in 
the previous diagram, there is a unique morphism $s_i:V\rightarrow Y_i\times_X V$ such that $(p'\circ p'_i)\circ s_i=id_V$, i.e. 
$s_i$ is a section of $p'\circ p'_i:Y_i\times_V\rightarrow V$, i.e. $f_i$ has a rational section.

On the other hand, assume that $f_i$ have rational sections, for some $i\in I$ % Using the induction on $n=\#I$, we show that $f$ has a rational section. For $n=1$, it is automatic, assume the statement is satisfied for $k<n$, and we prove it below for $n$:\\ 
Let $Y'=\displaystyle\bigcup_{i=0}^{n-1} Y_i$, then the induced morphism $f':=f|_{Y'}:Y'\rightarrow X$ admits a rational section 
over some $\xymatrix{U'\ar@{^(->}[r]|{\circ}&X}$. 
Let $f_i:Y_n\rightarrow X$ admits a rational section $s_i$ over some dense $\xymatrix{U_i\ar@{^(->}[r]|{\circ}&X}$. Consider the 
Cartesian diagram:
\[\xymatrix{
&Y\ar[r]^f&X\\
Y_i\ar@{^(->}[ur]|{\circ}&Y\times_X U_i\ar[r]^p\ar@{^(->}[u]|{\circ}&U_i\ar@{^(->}[u]|{\circ}\ar@/^/[lld]^s\\
Y_i\times_X U_i\ar@{^(->}[u]|{\circ}\ar@{^(->}[ru]|{\circ}_{p_i}&&
}
.\]
Then $(p\circ p_i)\circ s_i=id_U$, hence, $p\circ (p_i\circ s_i)=id_U$, i.e. $p_i\circ s_i$ is a section for $p$. Hence, $f$ has a 
rational section over $U_i$.
\tcr{Double check the proof}.

\end{proof}


\begin{lemma}
Let $f:Y\rightarrow X$ be an morphism of schemes, $X$ Noetherian, then $f$ has a rational section iff every $f_i:Y\times_X X_i
\rightarrow X$ has a rational section for every $i\in I$, where $\{X_i\}_{i\in I}$ are the irreducible components of $X$.
\end{lemma}
\begin{proof}
Assume $f$ has a rational section $s$ over $\xymatrix{U\ar@{^(->}[r]|{\circ}&X}$, $p$ is the projection $p:Y\times_X U\rightarrow 
U$, then $p\circ s=id_U$. Let  $\xymatrix{X_i\ar@{^(->}[r]|{\circ}&X}$ be an irreducible component of $X$, we have the Cartesian 
diagram:
\[\xymatrix{
&Y\ar[rr]^f&&X\\
Y\times_X X_i\ar@{^(->}[ur]|{\circ}^{f'_i}\ar[rr]_{f_i}&&X_i\ar@{^(->}[ur]|{\circ}_i\\
&&Y\times_X U\ar@{^(..>}[uul]|{\circ}^{p'}\ar@{^(..>}[rr]|{\circ}_{p}&&U\ar@{..>}@/_/[ll]_s\ar@{^(->}[uul]|{\circ}_{i_U}\\
&Y\times_X(X_i\times_X U)\ar@{^(->}[uul]|{\circ}_{\pi''}\ar@{^(..>}[ur]|{\circ}^{\pi'}\ar[rr]_{\pi}&&X_i\times_X U \ar@{^(->}
[uul]|{\circ}_{j}\ar@{^(->}[ur]|{\circ}_{j_U}\ar@{..>}[llluu]|{s'}\ar@/_/@{..>}[ll]_{s_i}
}
.\]
Now, consider the composition $g=p'\circ s \circ j_U:X_i\times_X U\rightarrow Y$, and the open embedding $j_i:X_i\times_X U
\hookrightarrow U$, then, since $p\circ s=id_U$, tracking the diagram, we find that the extension of these two morphism to $X$ 
coincide, i.e. $i\circ j=f\circ g$ hence, $\exists!s':X_i\times_X U\rightarrow Y\times_X X_i$ such that:
\[j=f_i\circ s' \text{ and } g=f'_i\circ s'
\]
Now, considering $s'$ and $l=s\circ j_U:X_i\times_X U\rightarrow Y\times_X U$, $g=p'\circ l$, then having $f'_i\circ s'=g=p'\circ 
l$ implies that $\exists! s_i:X_i\times_X U\rightarrow Y\times_X(X_i\times_X U)$ such that:
\[
s'=\pi''\circ s_i \text{ and } s\circ j_U=l= \pi' \circ s_i.
\]
$s_i$ is a section of $\pi$, that having $s\circ J_U= \pi' \circ s_i$ tracking the bottom diagram, we find:
\[
j_U\circ \pi \circ s_i=p\circ \pi'\circ s_i=p\circ s\circ j_U=id_U\circ j_U=j_U.
\]
$j_U$ in an open embedding, hence a monomorphis in $\Sch/S$, then $\pi\circ s_i=id_{X_i\times U}$. 
Notice that $(Y\times_X X_i)\times_{X_i}(X_i\times_X U) \cong Y \times_X(X_i\times_X U)$, and since $X_i$ is irreducible, then $|
X_i\times U|$ is dense in $X_i$, hence $f_i$ has rational section $s_i$ over $|X_i\times U$.




%On the other hand, \tcr{$X$ is Noetherian, then $I$ is finite.} Assume $f_i$'s has rational sections $s_i$ over $U_i$ for all $i\in I$. Then, by the universal property of co-products, we have the below commutative diagram:
%\[\xymatrix{&Y\ar[rr]|f&&X&\\\displaystyle\coprod_{i\in I}Y_i\ar@{..>}[rrrr]|{\displaystyle\coprod_{i\in I}f_i}\ar@{..>}[ru]|{q'}  &&&&\displaystyle\coprod_{i\in I}X_i\ar[lu]|q\\&Y_i \ar[rr]|{f_i}\ar@{^(->}[uu]|{f'_i}\ar@{_(->}[lu]|{j'_i}&&X_i\ar@{^(->}[uu]|i\ar@{^(->}[ru]|{j_i}&\\\displaystyle\coprod_{i\in I}V_i \ar@{^(->}[uu]^{\displaystyle\coprod_{i\in I}p'_i	}\ar@{..>}[rrrr]_{\displaystyle\coprod_{i\in I}p_i}&&&&\displaystyle\coprod_{i\in I} U_i \ar@{^(->}[uu]_{\displaystyle\coprod_{i\in I} i'}\ar@{red}@{-->}@/_/[llll]_{\displaystyle\coprod_{i\in I}s_i}\\&V_i \ar@{^(->}[uu]|{p'_i}\ar@{_(->}[ul]|{k'_i}\ar[rr]|{p_i}&& U_i \ar@/_/[ll]_{s_i} \ar@{^(->}[uu]|{i'}\ar@{^(->}[ur]|{k_i}&}\]
%where $Y_i:=Y\times_X X_i$ and $V_i:=Y\times_X U_i$.
%Note that since product of schemes is distributive over coproducts, then \tcb{$\displaystyle\coprod_{i\in I}Y_i\cong Y\times_X (\displaystyle\coprod_{i\in I}X_i)$ and$\displaystyle\coprod_{i\in I}V_i\cong Y\times_X (\displaystyle\coprod_{i\in I}U_i)$.}

%\noindent We notice that $s:=\displaystyle\coprod_{i\in I}s_i$ is a section of $p:=\displaystyle\coprod_{i\in I}p_i$ that $\forall i\in I$:
%\[p\circ s\circ k_j=p\circ k'_i\circ s_i=k_i\circ p_i\circ s_i=k_i\]
%by the universal property of $\displaystyle\coprod_{i\in I}U_i$ we have $p\circ s=id_{\displaystyle\coprod_{i\in I}U_i}$.

%Then, consider the subdiagram


On the other hand, \tcr{$X$ is Noetherian, then $I$ is finite.} Assume $f_i$'s has rational sections $s_i$ over $U_i$ for all $i\in 
I$. 
\[
\xymatrix{
&Y\ar[rr]|f&&X&\\
&Y_i \ar[rr]|{f_i}\ar@{^(->}[u]|{f'_i}&&X_i\ar@{^(->}[u]|i&\\
\displaystyle\bigcup_{i\in I}V_i \ar@{^(->}[uur]^{ \overline{p'}}\ar@{..>}[rrrr]_{\overline{p}}&&&&\displaystyle\bigcup_{i\in I} 
U_i \ar@{^(->}[uul]_{\overline{i'}}\ar@{red}@{-->}@/_/[llll]_{\overline{s}}\\
&V_i \ar@{^(->}[uu]|{p'_i}\ar@{_(->}[ul]|{k'_i}\ar[rr]|{p_i}&& U_i \ar@/_/[ll]_{s_i} \ar@{^(->}[uu]|{i'}\ar@{^(->}[ur]|{k_i}&\\
&V \ar@{^(->}[u]|{p'}\ar@{_(->}[uul]|{k'}\ar[rr]|{p}&& U \ar@{-->}@/_/[ll]_{s} \ar@{^(->}[u]|{j}\ar@{^(->}[uur]|{k}&}
\tag{*}\label{RatSecDiagram}
\]
where $Y_i:=Y\times_X X_i$, $V_i:=Y\times_X U_i$, $U$ an open subschme of $U_i$, $V:=Y\times_X U$, $i,i',j,k_i,k'_i,k,k'$ are open 
embeddings, and $f_i, p_i, \text{ and } p$ are the canonical morphism induced by the fiber product.

\noindent Note that
\[
f\circ (f'_i\circ p'_i\circ s_i\circ j)=(i\circ i' \circ p_i)\circ (\circ s_i\circ j)=i\circ i' \circ j
\]
Hence, there exist unique such $s$, induced by the universal property of fiber product, that makes the below sub-diagram commute:
\[
\xymatrix{
&&U\ar[dr]^{i\circ i'\circ j}\\
U\ar@/^/[urr]^{id}\ar@/_/[drr]_{f'_i\circ p'_i\circ s_i\circ j}\ar@{-->}[r]^s&Y\times_X U  \ar[ur]|{p}\ar[dr]^{f'_i\circ p'_i\circ 
p'}&&X\\
&&Y\ar[ru]_f&
}
\]
the commutativity of the sub-diagram implies $s\circ p=id_U$, i,e, $s$ is a section over $U$, also, it implies that $(f'_i\circ 
p'_i)\circ (s_i\circ j)=(f'_i\circ p'_i)\circ (p_i\circ s)$, since $(f'_i\circ p'_i)$ is an open embedding, we find that $s_i\circ 
j=p_i\circ s$, i.e. the front side of \eqref{RatSecDiagram} commutes.

\noindent Also, the side diagrams of \eqref{RatSecDiagram} commutes, by lemma \ref{SchOpUnion}.

\noindent Since the front diagram, commutes for such arbitrary open $U$, then by the universal property of $\displaystyle\bigcup_
{i\in I} U_i$, there exist unique such $\overline{s}$ that satisfies:
\noindent \[\overline{s}\circ k=k'\circ s \text{ and } \overline{s}\circ k_i=k'_i\circ s_i\]
\noindent Similarly, using \ref{SchOpSubFiber}, then then by the universal property of $\displaystyle\bigcup_{i\in I} V_i$, there 
exist unique such $\overline{p}$ that satisfies:
\noindent \[k\circ p=\overline{p} \circ k'  \text{ and } k_i \circ p_i=\overline{p} \circ k'_i\]

\noindent Notice that for any such $k$
\[
(\overline{p}\circ \overline{s})\circ k=\overline{p}\circ (k'\circ s)=k\circ (p\circ s)=k
\]
\noindent Then by the universal property of $\displaystyle\bigcup_{i\in I}U_i$ we find $\overline{p}\circ \overline{s}=id_
{\displaystyle\bigcup_{i\in I}U_i}$.

Since $I$ is finite, then by \ref{SchDisFberUnion}, $\displaystyle\bigcup{i\in I} V_i\cong Y\times_X \displaystyle\bigcup{i\in I} 
U_i$, hence $\overline{s}$ is a section for $\overline{p}$. It is seen readily that $|\displaystyle\bigcup{i\in I} U_i|$ is dense 
in $X$, hence, $f$ has rational section over $\displaystyle\bigcup{i\in I} U_i$.
\end{proof}
\begin{lemma}\label{RationalSections}
Let $f:Y\rightarrow X$ be a morphism between \tcb{Noetherian} \tcb{irreducible schemes}, its \tcr{generic} point $x\in 
X$, then rational sections $s$ of $f$ over $U$ are in bijection with the set of pairs $(y,s_x)$, where $f(y)=x$ and $s_x:\bcO_
{Y,y}\rightarrow \bcO_{X,x}$ is a local section of $f_y:\bcO_{X,x}\rightarrow \bcO_{Y,y}$.
\end{lemma}
\begin{proof}
Let $s$ be a rational section of $f$ over $U$
\[
\xymatrix{
Y\ar[r]^f&X\\
Y\times_X U\ar[r]_p\ar@{^(->}[u]|{\circ}^j& U\ar@/_/[l]_s\ar@{^(->}[u]|{\circ}_i
}
\]
$x\in |U|$, otherwise it cannot be generic for the irreducible $X$. We define $y=(j\circ s)(x)$, then $f(y)=((f\circ j)\circ s)(x)
=(i\circ (p\circ s))x=i(x)=x$, and since the composition $p\circ s=id_U$, and that the embeddings of the level of local rings is 
the identity, then $s_x\circ f_y=id_{\bcO_{X,x}}$ (We identify $\bcO_{U,x}=\bcO_{X,x}$ and $\bcO_{Y,y}=\bcO_{Y\times_X U,y}$).
%\here

\noindent On the other hand, let $y\in Y$ over $x$ such that, and assume the existence of the local section $s_x$ of $f_y$. Since, $Y$ is Noetherian, then $f$ is of finite type. \tcr{type from EGA, and Joe's explanation}, the irreducibility implies that it is dense.
\end{proof}
\begin{lemma}\label{etaleLift}
Let $f:Y\rightarrow X$ be an \'etale morphism of schemes, and completely decomposed at $x\in X$, then for any henselian ring $B$, 
the morphism
\[
\hom(f^{\#}_{\rf(x)},Id_B):\hom(\bcO_{Y,y},B)\rightarrow\hom(\bcO_{X,x},B)
\]
is bijection, where $\rf(x)\cong \rf(y)$.
\end{lemma}
\begin{proof}
\tcr{EGA4 18.6.2}
\end{proof}
\begin{lemma}
Let $x$ be a generic point of $X$, then $\bcO_{X,x}$ is Artinian local ring.
\end{lemma}
\begin{proof}
\tcr{EGA1 7.1.5}
\end{proof}
\begin{lemma}
Artinian local rings are complete.
\end{lemma}
\begin{proof}
\tcr{Milne}
\end{proof}

\begin{lemma}\label{RationalSectionsGeneric}
Let $f:Y\rightarrow X$ be an \'etale morphism between Noetherian schemes, and completely decomposed at every \tcb{generic} point of 
$X$, then $f$ has a rational section.
\end{lemma}
\begin{proof}
Let $\{X_i\}_{i\in I}$ be the irreducible components of $X$, $x_i$ the generic points of $X_i$, $Y_i:=Y\times_X X_i$, and $f_i:Y_i
\rightarrow X_i$ the pull-back of $f$ along the open embedding $X_i\hookrightarrow X$. Since, \'etale morphisms, and completly 
decomposable morphisms are stable under base changes, and that the generic points of $X$ are by definition the generic points of 
its irreducible components, then $f_i$ is \'etale and completly decomposed at $x_i$. Let $y_i\in Y_i$ be over $x_i$ such that 
$(f_i^{\#})_{\rf(y_i)}:\rf(x_i)\rightarrow \rf(y_i)$ is an isomorphism. Then, since $f_i$ is \'etale, and using \ref{etaleLift}, 
there exist a local lift
\[
\xymatrix{
\bcO_{X_i,x_i}\ar[r]\ar[d]&\bcO_{X_i,x_i}^h\\
\bcO_{Y_i,y_i}\ar@{-->}[ru]
}
\]
Since $\bcO_{X_i,x_i}$ is complete, then $\bcO_{X_i,x_i}\cong \bcO_{X_i,x_i}^{\#}$ is an isomorphism, then there exist a \tcb{local 
section} $s_i:\bcO_{Y_i,y_i}\rightarrow \bcO_{X_i,x_i}$, which in turn implies the existence of a rational section of $f_i$ over 
$U_i\subseteq X_i$, hence there exist a rational section of $f$ over $U=\displaystyle\bigcup_{i\in I}U_i$.
\end{proof}
%\here

\begin{lemma}\label{NisSplitting}
Every Nisnevich covering of a scheme $X$, \tcr{over $S$}, admits a splitting sequence.
\end{lemma}
\begin{proof}
Let $\bcU=\{f_i:X_i\rightarrow X\}$ be a Nisnevich covering of $X$, \tcr{over $S$}, let $Y=\displaystyle\coprod_{i\in I}X_i$ and 
$f=\displaystyle\coprod_{i\in I}f_i$ then, by lemma \ref{Nis_Coprod_Mor}, $\{f:Y\rightarrow X\}$ is a Nisnevich covering of $X$.

\tcb{Since $Y, X$ are over $S$, then they are Noetherian (we just need quasi compactness here!)}. Put $Z_0=X$, and let $x_0$ be a generic point of $Z_0$, then, using lemma \ref{RationalSectionsGeneric}, there exist $\xymatrix{x_0\in U_0\ar@{^(->}[r]|{\circ}&Z_0}$ dense, 
and there exist a section $s_0:U_0\rightarrow Y\times_X U_0$ of the canonical projection $p_0: Y\times_X U_0\rightarrow U_0$. Let $Z_1=(Z_0-U_0)_{red}$, and consider $f_1:Y\times_X Z_1\rightarrow Z_1$ be the canonical projection, then it is a Nisnevich covering for $Z_1$ because Nisnevich covering is stable under base change. Therefore, we can repeat the above construction, so we get the sequence of closed sub-schemes of $X$:
\[
... \nhookrightarrow Z_{n+1}\nhookrightarrow Z_n \nhookrightarrow Z_{n-1} \nhookrightarrow ... \nhookrightarrow Z_0=X
\]
where all below projections $p_{j}:(Z_j-Z_{j+1})\times_X Y\rightarrow Z_j-Z_{j+1}$ have \hyperref
[Section]{$S\!-\!$sections} $s_{j}:Z_j-Z_{j+1}\rightarrow (Z_j-Z_{j+1})\times_X Y$ for $0\leq j\leq n$:
\[
\xymatrix{
(Z_j-Z_{j+1})\times_X Y\ar@{^(->}[rr]\ar[d]^{p_{j}}&&Z_j\times_X Y\ar@{^(->}[rr]\ar[d]&&Y\ar[d]^{f}\\
U_j=Z_j-Z_{j+1}\ar@{^(->}[rr]|{\bigcirc}\ar@/^/@{-->}[u]^{s_{j}}&&Z_j\ar@{^(->}[rr]|{/}&&X
}
\]
Since $U_j$ is dense in $Z_j$, then $Z_j\neq Z_{j+1}$ where $Z_j\neq \emptyset$, but since $X$ is Noetherian, there exists $n$ such that $Z_n=\emptyset$. Hence, $\bcU$ admits splitting sequence \tcb{adjust in line with the main definition of splitting sequences}.
\end{proof}
\begin{lemma}
Let $\bcP$ be a pre-sheaf on $\Sch/S_{Nis}$, $\bcU=\{f_i:X_i\rightarrow X\}$ a Nisnevich covering, $Y=\displaystyle\coprod_{i\in I}X_i$, and $f=\displaystyle\coprod_{i\in I}X_i$, then the exactness of the below sequence:
\[
\bcP(X)\rightarrow \displaystyle\prod_{i\in I}\bcP(X_i)\rightrightarrows \displaystyle\prod_{i,j\in I}\bcP(X_i\times_X X_j)
\]
is equivalent to the exactness of:
\[
\bcP(X)\rightarrow \bcP(Y)\rightrightarrows \bcP(Y\times_X Y)
\]
\end{lemma}
\begin{proof}
\tcr{type}
\end{proof}
\begin{lemma}\label{SectionSheaf}
Let $\bcP$ be a pre-sheaf on $\Sch/S$, $f:Y\rightarrow X$ is morphism of schemes, if $f$ has a section, then the following sequence is split exact.
\[
\bcP(X)\rightarrow \bcP(Y)\rightrightarrows \bcP(Y\times_X Y)
\]
\end{lemma}
\begin{proof}
\tcr{type use that $I=s\circ f$ is idempotent, and use the decomposition $id_Y=(id_Y-I)+I$}.
\end{proof}

\begin{proposition}
A pre-sheaf $\bcP$ on $\Sch/S_{Nis}$ is a sheaf iff for any elementary distinguished square, as in Def \ref{ElemDisSq}, the 
following square of sets is Cartesian:
\[\xymatrix{
\bcP(X)\ar[r]\ar[d]&\bcP(U)\ar[d]\\
\bcP(Y)\ar[r]&\bcP(U\times_X Y)
}\]
\end{proposition}
\begin{proof}
%%\here

Let $\bcP$ be a pre-sheaf on $\Sch/S_{Nis}$ that satisfies the conditions of the proposition, and let $\bcU=\{f_i:X_i\rightarrow X\}_{i\in I}$ be a Nisnevich covering of $X$, over $S$, and let $Y=\displaystyle\coprod_{i\in I}X_i$, and $f=\displaystyle\coprod_{i\in I}X_i$. Then, $\bcU$ admits a splitting sequence, as seen in lemma \ref{NisSplitting}. Let $n$ be the smallest natural number such that
\[
\emptyset= Z_{n+1}\nhookrightarrow Z_n \nhookrightarrow Z_{n-1} \nhookrightarrow ... \nhookrightarrow Z_0=X
\]
is a splitting sequence of $\bcU$. The exactness of \[\bcP(X)\rightarrow \bcP(Y)\rightrightarrows \bcP(Y\times_X Y)\tag{*}\label{NisSheaf}\] could be proven by induction on $n$:\\
For $n=0$, $f$ has a section $s$, then by lemma \ref{SectionSheaf} the sequence \eqref{NisSheaf} is exact.\\
For $n>0$, \tcr{finish typing here}

\end{proof}
\tcr{This proposition means that...}
\noindent Nisnevich topology on $\Sm/S$ is stronger than Zariski topology, that every Zariski covering is a Nisnevich covering. 
But, it is weaker than \'etale topology, that by definition every Nisnevich covering is a \'etale covering, and some \'etale 
coverings are not Nisnevich as shown in the following example:
\begin{example}
Let $\kk$ be a field, $\alpha\in \kk^{\times}$, and consider the family $\bcU:\{(-)^2:\AF_{\kk}-\{0\}\rightarrow \AF_{\kk}, f_
{\alpha}:\AF_{\kk}-\{[(x-\alpha)]\}\hookrightarrow \AF_{\kk}\}$, where $f_{\alpha}$ is the open embedding, and $(-)^2$ is the 
morphism of affine schemes induced by $\varphi:\kk[x]\rightarrow \kk[x]_x$ that sends $x$ to $x^2$.Then, $\bcU$ is an \'etale 
covering for any $\alpha\in \kk^{\times}$, whereas it is Nisnevich only for $\alpha\in (\kk^{\times})^2$.
\end{example}
\begin{proof}
$\forall \beta \in \kk \varphi^{-1}((x-\beta))=(x-\beta^2) $ that:\\
$\forall g(x)\in (x-\beta^2), \exists h(x)\in \kk[x]_x$ such that $g(x)=(x-\beta^)h(x)$, hence $\varphi(g(x))=(x^2-\beta^2)h^2(x)=
(x-\beta)(x+\beta)h^2(x)\in (x-\beta)$.\\
Then, $(x-\beta^2)\subseteq \varphi^{-1}((x-\beta))$.  $(x-\beta^2)$ is the maximum ideal, hence $\varphi^{-1}((x-\beta))=(x-
\beta^2)$. Therefore, if $\exists \alpha \notin \kk^2$ then $(-)^2$ is not surjective.
\tcr{it seems that the issue is with the subjectivity which make fail to be \'etale as well!!}
\end{proof}
Alternative example is that an infinite family of \'etale morphisms.
\noindent \tcb{Open immersion is locally of finite type!}

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